表:
UserId, Value, Date.
我想获得UserId,为每个UserId的最大值(日期)的值。也就是说,具有最新日期的每个UserId的值。有没有一种方法可以在SQL中简单地做到这一点?(最好是Oracle)
更新:为任何歧义道歉:我需要得到所有的用户id。但是对于每个UserId,只有该用户拥有最新日期的行。
表:
UserId, Value, Date.
我想获得UserId,为每个UserId的最大值(日期)的值。也就是说,具有最新日期的每个UserId的值。有没有一种方法可以在SQL中简单地做到这一点?(最好是Oracle)
更新:为任何歧义道歉:我需要得到所有的用户id。但是对于每个UserId,只有该用户拥有最新日期的行。
当前回答
SELECT a.userid,a.values1,b.mm
FROM table_name a,(SELECT userid,Max(date1)AS mm FROM table_name GROUP BY userid) b
WHERE a.userid=b.userid AND a.DATE1=b.mm;
其他回答
select userid, value, date
from thetable t1 ,
( select t2.userid, max(t2.date) date2
from thetable t2
group by t2.userid ) t3
where t3.userid t1.userid and
t3.date2 = t1.date
恕我直言,这是可行的。HTH
答案是Oracle。这里有一个更复杂的SQL回答:
谁的整体作业成绩最好(作业点数最多)?
SELECT FIRST, LAST, SUM(POINTS) AS TOTAL
FROM STUDENTS S, RESULTS R
WHERE S.SID = R.SID AND R.CAT = 'H'
GROUP BY S.SID, FIRST, LAST
HAVING SUM(POINTS) >= ALL (SELECT SUM (POINTS)
FROM RESULTS
WHERE CAT = 'H'
GROUP BY SID)
还有一个更难的例子,需要一些解释,我没有时间了
给出2008年最受欢迎的书(ISBN和书名),即2008年最常被借阅的书。
SELECT X.ISBN, X.title, X.loans
FROM (SELECT Book.ISBN, Book.title, count(Loan.dateTimeOut) AS loans
FROM CatalogEntry Book
LEFT JOIN BookOnShelf Copy
ON Book.bookId = Copy.bookId
LEFT JOIN (SELECT * FROM Loan WHERE YEAR(Loan.dateTimeOut) = 2008) Loan
ON Copy.copyId = Loan.copyId
GROUP BY Book.title) X
HAVING loans >= ALL (SELECT count(Loan.dateTimeOut) AS loans
FROM CatalogEntry Book
LEFT JOIN BookOnShelf Copy
ON Book.bookId = Copy.bookId
LEFT JOIN (SELECT * FROM Loan WHERE YEAR(Loan.dateTimeOut) = 2008) Loan
ON Copy.copyId = Loan.copyId
GROUP BY Book.title);
希望这能对(任何人)有所帮助。:)
问候 古斯
首先,我看错了问题,下面是一个完整的例子,结果是正确的:
CREATE TABLE table_name (id int, the_value varchar(2), the_date datetime);
INSERT INTO table_name (id,the_value,the_date) VALUES(1 ,'a','1/1/2000');
INSERT INTO table_name (id,the_value,the_date) VALUES(1 ,'b','2/2/2002');
INSERT INTO table_name (id,the_value,the_date) VALUES(2 ,'c','1/1/2000');
INSERT INTO table_name (id,the_value,the_date) VALUES(2 ,'d','3/3/2003');
INSERT INTO table_name (id,the_value,the_date) VALUES(2 ,'e','3/3/2003');
--
select id, the_value
from table_name u1
where the_date = (select max(the_date)
from table_name u2
where u1.id = u2.id)
--
id the_value
----------- ---------
2 d
2 e
1 b
(3 row(s) affected)
如果你在使用Postgres,你可以使用array_agg像
SELECT userid,MAX(adate),(array_agg(value ORDER BY adate DESC))[1] as value
FROM YOURTABLE
GROUP BY userid
我不熟悉甲骨文。这是我想到的
SELECT
userid,
MAX(adate),
SUBSTR(
(LISTAGG(value, ',') WITHIN GROUP (ORDER BY adate DESC)),
0,
INSTR((LISTAGG(value, ',') WITHIN GROUP (ORDER BY adate DESC)), ',')-1
) as value
FROM YOURTABLE
GROUP BY userid
两个查询返回的结果都与接受的答案相同。看到SQLFiddles:
接受的答案 我对Postgres的解决方案 我对甲骨文的解决方案
我想是这样的。(请原谅我的语法错误;在这一点上,我习惯使用HQL !)
编辑:也误解了问题!修正了查询…
SELECT UserId, Value
FROM Users AS user
WHERE Date = (
SELECT MAX(Date)
FROM Users AS maxtest
WHERE maxtest.UserId = user.UserId
)